[HDU] 4027 - Can you answer these queries? - 線段樹

Can you answer these queries?

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 20101    Accepted Submission(s): 4737


Problem Description
A lot of battleships of evil are arranged in a line before the battle. Our commander decides to use our secret weapon to eliminate the battleships. Each of the battleships can be marked a value of endurance. For every attack of our secret weapon, it could decrease the endurance of a consecutive part of battleships by make their endurance to the square root of it original value of endurance. During the series of attack of our secret weapon, the commander wants to evaluate the effect of the weapon, so he asks you for help.
You are asked to answer the queries that the sum of the endurance of a consecutive part of the battleship line.

Notice that the square root operation should be rounded down to integer.
 

Input
The input contains several test cases, terminated by EOF.
  For each test case, the first line contains a single integer N, denoting there are N battleships of evil in a line. (1 <= N <= 100000)
  The second line contains N integers Ei, indicating the endurance value of each battleship from the beginning of the line to the end. You can assume that the sum of all endurance value is less than 263.
  The next line contains an integer M, denoting the number of actions and queries. (1 <= M <= 100000)
  For the following M lines, each line contains three integers T, X and Y. The T=0 denoting the action of the secret weapon, which will decrease the endurance value of the battleships between the X-th and Y-th battleship, inclusive. The T=1 denoting the query of the commander which ask for the sum of the endurance value of the battleship between X-th and Y-th, inclusive.
 

Output
For each test case, print the case number at the first line. Then print one line for each query. And remember follow a blank line after each test case.
 

Sample Input
10 1 2 3 4 5 6 7 8 9 10 5 0 1 10 1 1 10 1 1 5 0 5 8 1 4 8
 

Sample Output
Case #1: 19 7 6
 

Source

今天航神 小太陽和毛子一起去打 的比賽
有一道這個

他們沒我做不出來啊哈哈哈哈哈哈哈哈哈哈哈哈哈哈哈哈哈(假裝只有我做得出來)

區間修改  每個元素開根號向下取整
區間查詢求和

數據範圍RT


恩一個數字最多根號 6 次就變成 1 了就沒必要開根號

所以放心大膽的把區間更新變成單點更新
用線段樹去維護

然後線段樹維護一下每個節點包含的區間是不是全都是 1 
如果是就不用往下更新了
別的照常

這題輸入數據是 LL 爲啥用 int 會被 T 啊
HDU的評測姬這麼傲嬌的麼

好像之前沒寫過區間修改換單點修改

剛開始有點寫壞了

不過還是A了
#include <bits/stdc++.h>
#define LL long long
using namespace std;

const int N = 100010;

struct Tree
{
    int l;
    int r;
    LL sum;
    bool o;
};

Tree tree[N * 4];
LL ma[N];
int n, m;

void up(int x)
{
    if(tree[x << 1].o && tree[x << 1 | 1].o){
        tree[x].o = true;
    }
    tree[x].sum = tree[x << 1].sum + tree[x << 1 | 1].sum;
}

void build(int x, int l, int r)
{
    tree[x].l = l;
    tree[x].r = r;
    tree[x].o = false;
    tree[x].sum = 0;

    if(l == r){
        tree[x].sum = ma[l];
        if(tree[x].sum == 1){
            tree[x].o = true;
        }
    }
    else{
        int mid;

        mid = l + ((r - l) >> 1);
        build(x << 1, l, mid);
        build(x << 1 | 1, mid + 1, r);
        up(x);
    }
}

void update(int x, int l, int r)
{
    int ll = tree[x].l;
    int rr = tree[x].r;


    if(tree[x].o || l > r){
        return;
    }
    if(ll == rr){
        tree[x].sum = (int)sqrt(tree[x].sum);
        if(tree[x].sum == 1){
            tree[x].o = true;
        }
    }
    else{
        int mid;

        mid = ll + ((rr - ll) >> 1);
        update(x << 1, l, min(mid, r));
        update(x << 1 | 1, max(mid + 1, l), r);
        up(x);
    }
}

LL quary(int x, int l, int r)
{
    int ll = tree[x].l;
    int rr = tree[x].r;

    if(l > r){
        return 0;
    }
    if(l == ll && rr == r){
        return tree[x].sum;
    }
    else{
        int mid;
        LL ans = 0;

        mid = ll + ((rr - ll) >> 1);
        ans += quary(x << 1, l, min(mid, r));
        ans += quary(x << 1 | 1, max(mid + 1, l), r);

        return ans;
    }
}

int main(int argc, char const *argv[])
{
    int nc = 1;

    while(scanf("%d", &n) == 1){
        printf("Case #%d:\n", nc ++);
        for(int i = 1; i <= n; i ++){
            scanf("%lld", &ma[i]);
        }
        build(1, 1, n);
        scanf("%d", &m);
        for(int i = 0; i < m; i ++){
            int t, l, r;

            scanf("%d%d%d", &t, &l, &r);
            if(l > r){
                swap(l, r);
            }
            if(t){
                printf("%lld\n", quary(1, l, r));
            }
            else{
                update(1, l, r);
            }
        }
        cout << endl;
    }


    return 0;
}


 
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